Acetaldehyde decomposes at 750 K: CH3CHO → CO + CH4. The reaction is first order in acetaldehyde and the half-life of the reaction is found to be 530 seconds. What is the rate constant for the reaction at this temperature?

k = 1.3 x 10⁻³ s⁻¹

Explanation:

For a first order reaction the integrated rate law is

Ln [A]t/[A]₀ = - kt

where [A] are the concentrations of acetaldehyde in this case, t is the time and k is the rate constant.

We are given the half life for the concentration of acetaldehyde to fall to one half its original value, thus

Ln [A]t/[A]₀ = Ln 1/2[A]₀/[A]₀= Ln 1/2 = - kt

- 0.693 = - k(530s) ⇒ k = 1.3 x 10⁻³ s⁻¹

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g